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The reagent (s) that can selectively pre...

The reagent (s) that can selectively precipitate `S^(2-)` from a mixture of `S^(2-)` and `SO_(4)^(2-)` in aqueous solution is (are)

A

`CuCl_(2)`

B

`BaCl_(2)`

C

`Pb (OOC CH_(3))_(2)`

D

`Na_(2) [Fe(CN)_(5) NO]`

Text Solution

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The correct Answer is:
To solve the problem of selectively precipitating \( S^{2-} \) from a mixture of \( S^{2-} \) and \( SO_4^{2-} \) in aqueous solution, we can follow these steps: ### Step 1: Identify the precipitating reagents We need to find reagents that will react with \( S^{2-} \) to form an insoluble precipitate while leaving \( SO_4^{2-} \) in solution. ### Step 2: Analyze the options 1. **Copper(II) chloride (\( CuCl_2 \))**: - When \( CuCl_2 \) is added to \( S^{2-} \), it forms copper sulfide (\( CuS \)), which is a black precipitate. - When \( CuCl_2 \) is added to \( SO_4^{2-} \), the resulting compound \( CuSO_4 \) is soluble in water. - Therefore, \( CuCl_2 \) selectively precipitates \( S^{2-} \) while leaving \( SO_4^{2-} \) in solution. 2. **Barium chloride (\( BaCl_2 \))**: - When \( BaCl_2 \) is added to \( S^{2-} \), it forms barium sulfide (\( BaS \)), which is a precipitate. - When \( BaCl_2 \) is added to \( SO_4^{2-} \), it forms barium sulfate (\( BaSO_4 \)), which is also a precipitate. - Since both ions precipitate, \( BaCl_2 \) cannot selectively precipitate \( S^{2-} \). 3. **Lead acetate (\( Pb(C_2H_3O_2)_2 \))**: - When \( Pb(C_2H_3O_2)_2 \) is added to \( S^{2-} \), it forms lead sulfide (\( PbS \)), which is a precipitate. - When \( Pb(C_2H_3O_2)_2 \) is added to \( SO_4^{2-} \), it forms lead sulfate (\( PbSO_4 \)), which is also a precipitate. - Thus, \( Pb(C_2H_3O_2)_2 \) cannot selectively precipitate \( S^{2-} \). 4. **Sodium compounds (e.g., \( Na_2SO_4 \))**: - Sodium compounds do not precipitate either \( S^{2-} \) or \( SO_4^{2-} \) as both remain soluble. - Therefore, sodium compounds cannot be used for selective precipitation. ### Step 3: Conclusion The only reagent that can selectively precipitate \( S^{2-} \) from the mixture is **Copper(II) chloride (\( CuCl_2 \))**. ### Final Answer The reagent that can selectively precipitate \( S^{2-} \) from a mixture of \( S^{2-} \) and \( SO_4^{2-} \) in aqueous solution is **\( CuCl_2 \)**. ---
To solve the problem of selectively precipitating \( S^{2-} \) from a mixture of \( S^{2-} \) and \( SO_4^{2-} \) in aqueous solution, we can follow these steps: ### Step 1: Identify the precipitating reagents We need to find reagents that will react with \( S^{2-} \) to form an insoluble precipitate while leaving \( SO_4^{2-} \) in solution. ### Step 2: Analyze the options 1. **Copper(II) chloride (\( CuCl_2 \))**: - When \( CuCl_2 \) is added to \( S^{2-} \), it forms copper sulfide (\( CuS \)), which is a black precipitate. ...
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